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motorcycle Legendary stuntman Evel Knievel blasts off in his custom rocket-powered Harley-Davidson Skycycle in an attempt to jump the Snake River Canyon in 1974. A parachute prematurely deployed and caused the craft to fall into the canyon, just short of the other side. Knievel survived. 3 OUTLINE CHAPTER Vectors and Two-Dimensional Motion In our discussion of one-dimensional motion in Chapter 2, we used the concept of vectors only to a limited extent. In our further study of motion, manipulating vector quantities will become increasingly important, so much of this chapter is devoted to vector techniques. Well then apply these mathematical tools to two-dimensional motion, especially that of projectiles, and to the understanding of relative motion. Bettmann/Corbis 3.1 3.2 3.3 3.4 3.5 Vectors and Their Properties Components of a Vector Displacement, Velocity, and Acceleration in Two Dimensions Motion in Two Dimensions Relative Velocity 3.1 VECTORS AND THEIR PROPERTIES Each of the physical quantities we will encounter in this book can be categorized as either a vector quantity or a scalar quantity. As noted in Chapter 2, a vector has both direction and magnitude (size). A scalar can be completely specied by its magnitude with appropriate units; it has no direction. An example of each kind of quantity is shown in Figure 3.1 (page 54). As described in Chapter 2, displacement, velocity, and acceleration are vector quantities. Temperature is an example of a scalar quantity. If the temperature of an object is 5 C, that information completely species the temperature of the object; no direction is required. Masses, time intervals, and volumes are scalars as well. Scalar quantities can be manipulated with the rules of ordinary arithmetic. Vectors can also be added and subtracted from each other, and multiplied, but there are a number of important differences, as will be seen in the following sections. When a vector quantity is handwritten, it is often represented with an arrow : over the letter ( A). As mentioned in Section 2.1, a vector quantity in this book 53 54 Chapter 3 Vectors and Two-Dimensional Motion : will be represented by boldface type with an arrow on top (for example, A ). The : magnitude of the vector A will be represented by italic type, as A. Italic type will also be used to represent scalars. Equality of Two Vectors. Two vectors A and B are equal if they have the same magnitude and the same direction. This property allows us to translate a vector parallel to itself in a diagram without affecting the vector. In fact, for most purposes, any vector can be moved parallel to itself without being affected. (See Fig. 3.2.) Adding Vectors. When two or more vectors are added, they must all have the same units. For example, it doesnt make sense to add a velocity vector, carrying units of meters per second, to a displacement vector, carrying units of meters. Scalars obey the same rule: It would be similarly meaningless to add temperatures to volumes or masses to time intervals. Vectors can be added geometrically or algebraically. (The latter is discussed : : at the end of the next section.) To add vector B to vector A geometrically, rst : draw A on a piece of graph paper to some scale, such as 1 cm 1 m, and so : that its direction is specied relative a coordinate system. Then draw vector B to : : the same scale with the tail of B starting at the tip of A , as in Active Figure 3.3a. : Vector B must be drawn along the direction that makes the proper angle rela: : : : tive vector A . The resultant vector R A B is the vector drawn from the tail : : of A to the tip of B . This procedure is known as the triangle method of addition. When two vectors are added, their sum is independent of the order of the : : : : addition: A B B A . This relationship can be seen from the geometric construction in Active Figure 3.3b, and is called the commutative law of addition. This same general approach can also be used to add more than two vectors, as : : : : is done in Figure 3.4 for four vectors. The resultant vector sum R A B C : D is the vector drawn from the tail of the rst vector to the tip of the last. Again, the order in which the vectors are added is unimportant. Negative of a Vector. The negative of the vector A is dened as the vector that : : : gives zero when added to A . This means that A and A have the same magnitude but opposite directions. : : : Figure 3.1 (a) The number of grapes in this bunch ripe for picking is one example of a scalar quantity. Can you think of other examples? (b) This helpful person pointing in the right direction tells us to travel ve blocks north to reach the courthouse. A vector is a physical quantity that must be specied by both magnitude and direction. TIP 3.1 Vector Addition Versus Scalar Addition A B C is very different from A B C. The rst is a vector sum, which must be handled graphically or with components, while the second is a simple arithmetic sum of numbers. : : : y A George Semple Mack Henley/Visuals Unlimited R= A+ B B B A R= +B O x A Figure 3.2 These four vectors are equal because they have equal lengths and point in the same direction. (a) (b) ACTIVE FIGURE 3.3 : : : : (a) When vector B is added to vector A, the vector sum R is: vector that runs from the tail of A the : : to the tip : B.: Here the resultant runs from the tail of B to the tip of A. These constructions prove of (b) : : that A B B A. Log into PhysicsNow at www.cp7e.com, and go to Active Figure 3.3 to vary A and B and see the effect on the resultant. : : 3.1 Vectors and Their Properties 55 : Vector subtraction is really a special case of vector addition. The geometric construction for subtracting two vectors is shown in Figure 3.5. Multiplying or Dividing a Vector by a Scalar. Multiplying or dividing a vector by a : scalar gives a vector. For example, if vector A is multiplied by the scalar number 3, : : the result, written 3 A, is a vector with a magnitude three times that of A and point: ing in the same direction. If we multiply vector A by the scalar 3, the result : : is 3 A , a vector with a magnitude three times that of A and pointing in the opposite direction (because of the negative sign). R =A +B A : B : A ( B) : [3.1] +C Subtracting Vectors. Vector subtraction makes use of the denition of the nega: : : tive of a vector. We dene the operation A B as the vector B added to the vec: tor A : D +D C B A Figure 3.4 A geometric construction for summing four vectors. : The resultant vector R is the vector that completes the polygon. Quick Quiz 3.1 The magnitudes of two vectors A and B are 12 units and 8 units, respectively. What are the largest and smallest possible values for the magnitude of the resultant : : : vector R B ? (a) 14.4 and 4; (b) 12 and 8; (c) 20 and 4; (d) none of these. A AB : : B A B Quick Quiz 3.2 If vector B is added to vector A , the resultant vector A B has magnitude A B : : when A and B are (a) perpendicular to each other; (b) oriented in the same direction; (c) oriented in opposite directions; (d) none of these answers. : : : : Figure 3.5 This construction : shows how to subtract vector B from : : vector A . The vector B has the : same magnitude as the vector B, but points in the opposite direction. EXAMPLE 3.1 Taking a Trip Goal Find the sum of two vectors by using a graph. N W S E y(km) 40 Problem A car travels 20.0 km due north and then 35.0 km in a direction 60 west of north, as in Figure 3.6. Using a graph, nd the magnitude and direction of a single vector that gives the net effect of the cars trip. This vector is called the cars resultant displacement. Strategy Draw a graph, and represent the displacement vectors as arrows. Graphically locate the vector resulting from the sum of the two displacement vectors. Measure its length and angle with respect to the vertical. Solution : : Let A represent the rst displacement vector, 20.0 km north, B the second displacement vector, extending west of north. Carefully graph the two vectors, drawing a : : : resultant vector R with its base touching the base of A and extending to the tip of B. Measure the length of this vector, which turns out to be about 48 km. The angle , measured with a protractor, is about 39 west of north. Remarks Notice that ordinary arithmetic doesnt work here: the correct answer of 48 km is not equal to 20.0 km 35.0 km 55.0 km! B 60.0 20 R 0 A x(km) 20 Figure 3.6 (Example 3.1) A graphical method for nding the resultant : : : displacement vector R A B. Exercise 3.1 Graphically determine the magnitude and direction of the displacement if a man walks 30.0 km 45 north of east and then walks due east 20.0 km. Answer 46 km, 27 north of east 56 y Chapter 3 Vectors and Two-Dimensional Motion 3.2 COMPONENTS OF A VECTOR Ay tan = Ax Ay A O Ax : x One method of adding vectors makes use of the projections of a vector along the axes of a rectangular coordinate system. These projections are called components. Any vector can be completely described by its components. : Consider a vector A in a rectangular coordinate system, as shown in Figure 3.7. : : : A can be expressed as the sum of two vectors: A x , parallel to the x-axis; and A y , parallel to the y -axis. Mathematically, : A : Ax : Ay : : Figure 3.7 Any vector A lying in the xy-plane can be represented by its rectangular components Ax and A y . TIP 3.2 x- and y-Components where A x and A y are the component vectors of A . The projection of A along the : : x-axis, A x , is called the x- component of A , and the projection of A along the : y-axis, A y , is called the y-component of A . These components can be either positive or negative numbers with units. From the denitions of sine and cosine, we : see that cos A x /A and sin A y /A, so the components of A are Ax Ay A cos A sin [3.2] : : Equation 3.2 for the x- and y-components of a vector associates cosine with the x-component and sine with the y-component, as in Figure 3.8a. This association is due solely to the fact that we chose to measure the angle with respect to the positive x-axis. If the angle were measured with respect to the y-axis, as in Figure 3.8b, the components would be given by A x A sin and A y A cos . These components form two sides of a right triangle having a hypotenuse with : magnitude A. It follows that A s magnitude and direction are related to its components through the Pythagorean theorem and the denition of the tangent: A Ax 2 Ay Ax Ay 2 [3.3] [3.4] TIP 3.3 Inverse Tangents on Calculators: Right Half the Time The inverse tangent function on calculators returns an angle between 90 and 90 . If the vector lies in the second or third quadrant, the angle, as measured from the positive x -axis, will be the angle returned by your calculator plus 180 . tan To solve for the angle , which is measured from the positive x-axis by convention, we can write Equation 3.4 in the form tan 1 Ay Ax y A Ay = A sin 0 A = A cos x (a) x y Ax = A sin Ay = A cos This formula gives the right answer only half the time! The inverse tangent function returns values only from 90 to 90 , so the answer in your calculator window will only be correct if the vector happens to lie in rst or fourth quadrant. If it lies in second or third quadrant, adding 180 to the number in the calculator window will always give the right answer. The angle in Equations 3.2 and 3.4 must be measured from the positive x-axis. Other choices of reference line are possible, but certain adjustments must then be made. (See Tip 3.2 and Figure 3.8.) If a coordinate system other than the one shown in Figure 3.7 is chosen, the components of the vector must be modied accordingly. In many applications its more convenient to express the components of a vector in a coordinate system having axes that are not horizontal and vertical, but are still perpendicular to each : other. Suppose a vector B makes an angle with the x -axis dened in Figure 3.9. : The rectangular components of B along the axes of the gure are given by Bx B cos and B y B sin , as in Equations 3.2. The magnitude and direc: tion of B are then obtained from expressions equivalent to Equations 3.3 and 3.4. y A x (b) By O B 0 x Figure 3.8 The angle need not always be dened from the positive x-axis. Bx Figure 3.9 The components of vector B in a tilted coordinate system. : 3.2 Components of a Vector 57 Quick Quiz 3.3 Figure 3.10 shows two vectors lying in the xy-plane. Determine the signs of the :: : : x- and y-components of A, B, and A B, and place your answers in the following table: Vector A : B : : y A x-component y-component x A : B B Figure 3.10 (Quick Quiz 3.3) EXAMPLE 3.2 Help Is on the Way! Goal Find vector components, given a magnitude and direction, and vice versa. Problem (a) Find the horizontal and vertical components of the 1.00 10 2 m displacement of a superhero who ies from the top of a tall building along the path shown in Figure 3.11a. (b) Suppose instead the superhero leaps in the other direction : along a displacement vector B , to the top of a agpole where 25.0 m and the displacement components are given by B x B y 10.0 m. Find the magnitude and direction of the displacement vector. Strategy (a) The triangle formed by the displacement and its components is shown in Figure 3.11b. Simple trigonometry gives the components relative the standard x-y coordinate system: A x A cos and A y A sin (Equations 3.2). Note that 30.0 , negative because its measured clockwise from the positive x- axis. (b) Apply Equations 3.3 and 3.4 to nd the magnitude and direction of the vector. y y 30.0 100 m x (a) Ax 30.0 A x Ay (b) Figure 3.11 (Example 3.2) Solution : (a) Find the vector components of A from its magnitude and direction. Use Equations 3.2 to nd the components of the : displacement vector A : (b) Find the magnitude and direction of the displace: ment vector B from its components. Compute the magnitude of B from the Pythagorean theorem: Calculate the direction of B using the inverse tangent, remembering to add 180 to the answer in your calculator window, because the vector lies in the second quadrant: Remarks In part (a), note that cos( ) cos ; however, sin( fact that displacement in the y-direction is downward. : : Ax Ay A cos A sin (1.00 (1.00 102 m) cos( 30.0 ) 102 m) sin( 30.0 ) 86.6 m 50.0 m B Bx 2 tan 158 ) 1 By 2 By Bx ( 25.0 m)2 10.0 25.0 (10.0 m)2 26.9 m tan 1 21.8 sin . The negative sign of A y reects the 58 Chapter 3 Vectors and Two-Dimensional Motion Exercise 3.2 (a) Suppose the superhero had own 150 m at a 120 angle with respect to the positive x- axis. Find the components of the displacement vector. (b) Suppose instead, the superhero had leaped with a displacement having an x-component of 32.5 m and a y-component of 24.3 m. Find the magnitude and direction of the displacement vector. Answers (a) Ax 75 m, Ay 130 m (b) 40.6 m, 36.8 Adding Vectors Algebraically The graphical method of adding vectors is valuable in understanding how vectors can be manipulated, but most of the time vectors are added algebraically in terms : : : of their components. Suppose R A B. Then the components of the resultant : vector R are given by Rx Ry Ax Ay Bx By [3.5a] [3.5b] So x-components are added only to x-components, and y-components only to y: components. The magnitude and direction of R can subsequently be found with Equations 3.3 and 3.4. Subtracting two vectors works the same way, because its a matter of adding the negative of one vector to another vector. You should make a rough sketch when adding or subtracting vectors, in order to get an approximate geometric solution as a check. INTERACTIVE EXAMPLE 3.3 Take a Hike Goal Add vectors algebraically and nd the resultant vector. Problem A hiker begins a trip by rst walking 25.0 km southeast from her base camp. On the second day she walks 40.0 km in a direction 60.0 north of east, at which point she discovers a forest rangers tower. (a) Determine the components of the hikers displacements in the rst and second days. (b) Determine the components of the hikers total displacement for the trip. (c) Find the magnitude and direction of the displacement from base camp. Strategy This is just an application of vector addition using components, Equations 3.5. We denote the displacement vectors on the rst and : : second days by A and B, respectively. Using the camp as the origin of the coordinates, we get the vectors shown in Figure 3.12a. After nding x- and y-components for each vector, we add them componentwise. Finally, we determine the magnitude and direction of the resultant vector : R , using the Pythagorean theorem and the inverse tangent function. Solution : (a) Find the components of A. Use Equations 3.2 to nd the components of A: : y(km) 20 10 Camp 10 20 (a) A R 45.0 20 30 40 B 60.0 50 W Tower N y(km) E S x(km) 30 20 10 O R R y = 16.9 km 40 x(km) 10 20 30 Rx = 37.7 km (b) Figure 3.12 (Example 3.3) (a) Hikers path and the resultant vector. (b) Components of the hikers total displacement from camp. Ax Ay A cos( 45.0 ) A sin( 45.0 ) B cos 60.0 B sin 60.0 (25.0 km)(0.707) (25.0 km)(0.707) 17.7 km 17.7 km Find the components of B: : Bx By (40.0 km)(0.500) (40.0 km)(0.866) 20.0 km 34.6 km 3.3 Displacement, Velocity, and Acceleration in Two Dimensions 59 (b) Find the components of the resultant vector, : : : R A B. To nd R x , add the x-components of A and B: To nd R y , add the y-components of A and B: (c) Find the magnitude and direction of R. Use the Pythagorean theorem to get the magnitude: Calculate the direction of R using the inverse tangent function: : : : : : : Rx Ry Ax Ay Bx By 17.7 km 17.7 km 20.0 km 34.6 km 37.7 km 16.9 km R R x 2 tan : Ry 2 (37.7 km)2 24.1 (16.9 km)2 41.3 km 1 16.9 km 37.7 km Remarks Figure 3.12b shows a sketch of the components of R and their directions in space. The magnitude and direction of the resultant can also be determined from such a sketch. Exercise 3.3 A cruise ship leaving port, travels 50.0 km 45.0 north of west and then 70.0 km at a heading 30.0 north of east. Find (a) the ships displacement vector and (b) the displacement vectors magnitude and direction. Register to View AnswerR x 25.3 km, R y 70.4 km (b) 74.8 km, 70.2 north of east Investigate this problem further by logging into PhysicsNow at www.cp7e.com and going to Interactive Example 3.3. 3.3 DISPLACEMENT, VELOCITY, AND ACCELERATION IN TWO DIMENSIONS In one-dimensional motion, as discussed in Chapter 2, the direction of a vector quantity such as a velocity or acceleration can be taken into account by specifying whether the quantity is positive or negative. The velocity of a rocket, for example, is positive if the rocket is going up and negative if its going down. This simple solution is no longer available in two or three dimensions. Instead, we must make full use of the vector concept. Consider an object moving through space as shown in Figure 3.13. When the object is at some point at time t i , its position is described by the position vector : r i , drawn from the origin to . When the object has moved to some other point r at time tf , its position vector is :f . From the vector diagram in Figure 3.13, the nal position vector is the sum of the initial position vector and the displacement :: : : r : rf ri r . From this relationship, we obtain the following one: An objects displacement is dened as the change in its position vector, or : y ti ri r tf Path of an object x rf O Figure 3.13 An object moving along some curved path between points and . The displacement r vector : is the difference in the : r r position vectors: : :f ri . r : rf : ri [3.6] Displacement vector SI unit: meter (m) We now present several generalizations of the denitions of velocity and acceleration given in Chapter 2. An objects average velocity during a time interval divided by t: : : t is its displacement v av r t [3.7] Average velocity SI unit: meter per second (m/s) 60 Chapter 3 Vectors and Two-Dimensional Motion Because the displacement is a vector quantity and the time interval is a scalar quantity, we conclude that the average velocity is a vector quantity directed along :. r Instantaneous velocity An objects instantaneous velocity : is the limit of its average velocity as v goes to zero: : : t v lim t:0 r t [3.8] SI unit: meter per second (m/s) The direction of the instantaneous velocity vector is along a line that is tangent to the objects path and in the direction of its motion. Average acceleration An objects average acceleration during a time interval t is the change in its velocity : divided by t, or v : : a av v t [3.9] SI unit: meter per second squared (m/s2) Instantaneous acceleration An objects instantaneous acceleration vector : is the limit of its average a acceleration vector as t goes to zero: : : a lim t:0 v t [3.10] SI unit: meter per second squared (m/s2) Its important to recognize that an object can accelerate in several ways. First, the magnitude of the velocity vector (the speed) may change with time. Second, the direction of the velocity vector may change with time, even though the speed is constant, as can happen along a curved path. Third, both the magnitude and the direction of the velocity vector may change at the same time. Quick Quiz 3.4 Which of the following objects cant be accelerating? (a) An object moving with a constant speed; (b) an object moving with a constant velocity; (c) an object moving along a curve. Quick Quiz 3.5 Consider the following controls in an automobile: gas pedal, brake, steering wheel. The controls in this list that cause an acceleration of the car are (a) all three controls, (b) the gas pedal and the brake, (c) only the brake, or (d) only the gas pedal. 3.4 MOTION IN TWO DIMENSIONS Projectile motion In Chapter 2, we studied objects moving along straight-line paths, such as the x -axis. In this chapter, we look at objects that move in both the x- and y-directions simultaneously under constant acceleration. An important special case of this twodimensional motion is called projectile motion. Anyone who has tossed any kind of object into the air has observed projectile motion. If the effects of air resistance and the rotation of Earth are neglected, the 3.4 y Motion in Two Dimensions 61 vy v0 v vy = 0 v 0x g v 0x vy v 0x v ACTIVE FIGURE 3.14 The parabolic trajectory of a particle that leaves the origin with a velocity of :0. Note that : changes with time. v v However, the x-component of the velocity, v x , remains constant in time. Also, v y 0 at the peak of the trajectory, but the acceleration is always equal to the free-fall acceleration and acts vertically downward. v 0y 0 v 0x v 0x 0 v 0y v x Log into PhysicsNow at www.cp7e.com, and go to Active Figure 3.14, where you can change the particles launch angle and initial speed. You can also observe the changing components of velocity along the trajectory of the projectile. path of a projectile in Earths gravity eld is curved in the shape of a parabola, as shown in Active Figure 3.14. The positive x-direction is horizontal and to the right, and the y-direction is vertical and positive upward. The most important experimental fact about projectile motion in two dimensions is that the horizontal and vertical motions are completely independent of each other. This means that motion in one direction has no effect on motion in the other direction. If a baseball is tossed in a parabolic path, as in Active Figure 3.14, the motion in the y-direction will look just like a ball tossed straight up under the inuence of gravity. Active Figure 3.15 shows the effect of various initial angles; note that complementary angles give the same horizontal range. In general, the equations of constant acceleration developed in Chapter 2 follow separately for both the x-direction and the y-direction. An important difference is that the initial velocity now has two components, not just one as in that chapter. We assume that at t 0, the projectile leaves the origin with an initial velocity :0 . If the velocity vector makes an angle 0 with the horizontal, where 0 is v called the projection angle, then from the denitions of the cosine and sine functions and Active Figure 3.14, we have v 0x v0 cos 0 TIP 3.4 Acceleration at the Highest Point The acceleration in the y-direction is not zero at the top of a projectiles trajectory. Only the y-component of the velocity is zero there. If the acceleration were zero, too, the projectile would never come down! and v 0y v 0 sin 0 where v 0x is the initial velocity (at t 0) in the x-direction and v 0y is the initial velocity in the y-direction. Now, Equations 2.6, 2.9, and 2.10 developed in Chapter 2 for motion with constant acceleration in one dimension carry over to the two-dimensional case; there is one set of three equations for each direction, with the initial velocities modied as just discussed. In the x-direction, with ax constant, we have y(m) 150 75 100 60 45 50 30 15 50 100 vi = 50 m/s ACTIVE FIGURE 3.15 A projectile launched from the origin with an initial speed of 50 m/s at various angles of projection. Note that complementary values of the initial angle result in the same value of R (the range of the projectile). 150 200 250 x(m) Log into PhysicsNow at www.cp7e.com, and go to Active Figure 3.15, where you can vary the projection angle to observe the effect on the trajectory and measure the ight time. 62 Chapter 3 Vectors and Two-Dimensional Motion vx x vx2 where v0x v0 cos 0. v 0x v 0x t v 0x2 axt 1 2 2 axt [3.11a] [3.11b] [3.11c] 2ax x In the y-direction, we have vy y v 0y v 0y t v 0y 2 a yt 1 2 2 ayt [3.12a] [3.12b] [3.12c] HIRB/Index Stock v y2 2a y y A water fountain. The individual water streams follow parabolic trajectories. The horizontal range and maximum height of a given stream of water depend on the elevation angle of that streams initial velocity as well as its initial speed. where v 0y v 0 sin 0 and a y is constant. The objects speed v can be calculated from the components of the velocity using the Pythagorean theorem: v vx2 tan 1 v y2 vy vx The angle that the velocity vector makes with the x -axis is given by This formula for , as previously stated, must be used with care, because the inverse tangent function returns values only between 90 and 90 . Adding 180 is necessary for vectors lying in the second or third quadrant. The kinematic equations are easily adapted and simplied for projectiles close to the surface of the Earth. In that case, assuming air friction is negligible, the acceleration in the x-direction is 0 (because air resistance is neglected). This means that a x 0, and the projectiles velocity component along the x -direction remains constant. If the initial value of the velocity component in the x -direction is v 0x v 0 cos 0, then this is also the value of vx at any later time, so vx v 0x v 0 cos 0 constant [3.13a] while the horizontal displacement is simply x v 0x t (v 0 cos 0)t [3.13b] g and [3.14a] For the motion in the y-direction, we make the substitution a y v 0y v 0 sin 0 in Equations 3.12, giving vy y vy 2 v 0 sin (v 0 sin (v0 sin 0 0)t 2 0) gt 1 2 2gt [3.14b] [3.14c] 2g y The important facts of projectile motion can be summarized as follows: 1. Provided air resistance is negligible, the horizontal component of the velocity vx remains constant because there is no horizontal component of acceleration. 2. The vertical component of the acceleration is equal to the free fall acceleration g. 3. The vertical component of the velocity v y and the displacement in the y-direction are identical to those of a freely falling body. 4. Projectile motion can be described as a superposition of two independent motions in the x - and y-directions. EXAMPLE 3.4 Projectile Motion with Diagrams Goal Approximate answers in projectile motion using a motion diagram. Problem A ball is thrown so that its initial vertical and horizontal components of velocity are 40 m/s and 20 m/s, respectively. Use a motion diagram to estimate the balls total time of ight and the distance it traverses before hitting the ground. 3.4 Motion in Two Dimensions 63 Strategy Use the diagram, estimating the acceleration of gravity as 10 m/s2. By symmetry, the ball goes up and comes back down to the ground at the same y-velocity as when it left, except with opposite sign. With this fact and the fact that the acceleration of gravity decreases the velocity in the y-direction by 10 m/s every second, we can nd the total time of ight and then the horizontal range. Solution In the motion diagram shown in Figure 3.16, the acceleration vectors are all the same, pointing downward with magnitude of nearly 10 m/s2. By symmetry, we know that the ball will hit the ground at the same speed in the y-direction Figure 3.16 (Example 3.4) Motion diagram for a projectile. as when it was thrown, so the velocity in the y-direction goes from 40 m/s to 40 m/s in steps of 10 m/s every second; hence, approximately 8 seconds elapse during the motion. The velocity vector constantly changes direction, but the horizontal velocity never changes, because the acceleration in the horizontal direction is zero. Therefore, the displacement of the ball in the x-direction is given by Equation 3.13b, x v0xt (20 m/s)(8 s) 160 m. Remarks This example emphasizes the independence of the x- and y-components in projectile motion problems. Exercise 3.4 Estimate the maximum height in this same problem. Answer 80 m Quick Quiz 3.6 Suppose you are carrying a ball and running at constant speed, and wish to throw the ball and catch it as it comes back down. Should you (a) throw the ball at an angle of about 45 above the horizontal and maintain the same speed, (b) throw the ball straight up in the air and slow down to catch it, or (c) throw the ball straight up in the air and maintain the same speed? Quick Quiz 3.7 As a projectile moves in its parabolic path, the velocity and acceleration vectors are perpendicular to each other (a) everywhere along the projectiles path, (b) at the peak of its path, (c) nowhere along its path, or (d) not enough information is given. Problem-Solving Strategy Projectile Motion 1. Select a coordinate system and sketch the path of the projectile, including initial and nal positions, velocities, and accelerations. 2. Resolve the initial velocity vector into x- and y-components. 3. Treat the horizontal motion and the vertical motion independently. 4. Follow the techniques for solving problems with constant velocity to analyze the horizontal motion of the projectile. 5. Follow the techniques for solving problems with constant acceleration to analyze the vertical motion of the projectile. 64 Chapter 3 Vectors and Two-Dimensional Motion EXAMPLE 3.5 Stranded Explorers Goal Solve a two-dimensional projectile motion problem in which an object has an initial horizontal velocity. Problem An Alaskan rescue plane drops a package of emergency rations to a stranded hiker, as shown in Figure 3.17. The plane is traveling horizontally at 40.0 m/s at a height of 1.00 10 2 m above the ground. (a) Where does the package strike the ground relative to the point at which it was released? (b) What are the horizontal and vertical components of the velocity of the package just before it hits the ground? Strategy Here, were just taking Equations 3.13 and 3.14, lling in known quantities, and solving for the remaining unknown quantities. Sketch the problem using a coordinate system as in Figure 3.17. In part (a), set the y -component of the displacement equations equal to 1.00 10 2 mthe ground level where the package lands and solve for the time it takes the package to reach the ground. Substitute this time into the displacement equation for the x-component to nd the range. In part (b), substitute the time found in part (a) into the velocity components. Notice that the initial velocity has only an x-component, which simplies the math. Solution (a) Find the range of the package. Use Equation 3.14b to nd the y-displacement: Substitute y 0 0 and v 0y 0, set y 1.00 10 2 m the nal vertical position of the package relative the airplane and solve for time: Use Equation 3.13b to nd the x-displacement: Substitute x 0 0, v 0x 40.0 m/s, and the time: x y t x y y y0 v 0y t 12 2 gt y 40.0 m/s x 100 m Figure 3.17 (Example 3.5) From the point of view of an observer on the ground, a package released from the rescue plane travels along the path shown. (4.90 m/s 2)t 2 4.52 s x x0 v 0x t 1.00 10 2 m (40.0 m/s)(4.52 s) 181 m (b) Find the components of the packages velocity at impact: Find the x-component of the velocity at the time of impact: Find the y-component of the velocity at the time of impact: vx vy v0 cos v 0 sin (40.0 m/s) cos 0 40.0 m/s gt 0 (9.80 m/s2)(4.52 s) 44.3 m/s Remarks Notice how motion in the x-direction and motion in the y-direction are handled separately. Exercise 3.5 A bartender slides a beer mug at 1.50 m/s towards a customer at the end of a frictionless bar that is 1.20 m tall. The customer makes a grab for the mug and misses, and the mug sails off the end of the bar. (a) How far away from the end of the bar does the mug hit the oor? (b) What are the speed and direction of the mug at impact? Answers (a) 0.742 m (b) 5.08 m/s, 72.8 3.4 Motion in Two Dimensions 65 EXAMPLE 3.6 The Long Jump Goal Solve a two-dimensional projectile motion problem involving an object starting and ending at the same height. Problem A long jumper (Fig. 3.18) leaves the ground at an angle of 20.0 to the horizontal and at a speed of 11.0 m/s. (a) How long does it take for him to reach maximum height? (b) What is the maximum height? (c) How far does he jump? (Assume that his motion is equivalent to that of a particle, disregarding the motion of his arms and legs.) (d) Find the maximum height he reaches using Equation 3.14c. A P P L I C AT I O N Long Jumping Solution (a) Find the time t max taken to reach maximum height. Set vy 0 in Equation 3.14b and solve for t max : Figure 3.18 (Example 3.6) Mike Powell, current holder of the world long-jump record of 8.95 m. vy t max v 0 sin v 0 sin g 0.384 s 0 0 g t max 0 (11.0 m/s)(sin 20.0 ) 9.80 m/s2 (b) Find the maximum height he reaches. Substitute the time t max into the equation for the y-displacement: y max y max y max (c) Find the horizontal distance he jumps. First nd the time for the jump, which is twice t max: Substitute this result into the equation for the x-displacement: (d) Use an alternate method to nd the maximum height. Use Equation 3.14c, solving for y: vy 2 v 0y 2 y 0 2g y vy 2 v0y2 2g ((11.0 m/s) sin 20.0 )2 2(9.80 m/s2 ) 0.722 m t x 2t max (v 0 cos 7.94 m 2(0.384 s) 0)t (v 0 sin 0)t max 1 2 2 g(t max) (11.0 m/s)(sin 20.0 )(0.384 s) 1 2 2 2 (9.80 m/s )(0.384 s) 0.722 m 0.768 s. (11.0 m/s)(cos 20.0 )(0.768 s) Substitute vy 0 at maximum height, and the fact that v 0y (11.0 m/s) sin 20.0 : y Mike Powell/Allsport/Getty Images Strategy Again, we take the projectile equations, ll in the known quantities, and solve for the unknowns. At the maximum height, the velocity in the y-direction is zero, so setting Equation 3.14a equal to zero and solving gives the time it takes him to reach his maximum height. By symmetry, given that his trajectory starts and ends at the same height, doubling this time gives the total time of the jump. 66 Chapter 3 Vectors and Two-Dimensional Motion Remarks Although modelling the long jumpers motion as that of a projectile is an oversimplication, the values obtained are reasonable. Exercise 3.6 A grasshopper jumps 1.00 m from rest, with an initial velocity at a 45.0 angle with respect to the horizontal. Find (a) the initial speed of the grasshopper and (b) the maximum height reached. Answers (a) 3.13 m/s (b) 0.250 m INTERACTIVE EXAMPLE 3.7 Thats Quite an Arm Goal Solve a two-dimensional kinematics problem with a nonhorizontal initial velocity, starting and ending at different heights. Problem A stone is thrown upward from the top of a building at an angle of 30.0 to the horizontal and with an initial speed of 20.0 m/s, as in Figure 3.19. The point of release is 45.0 m above the ground. (a) How long does it take for the stone to hit the ground? (b) Find the stones speed at impact. (c) Find the horizontal range of the stone. Strategy Choose coordinates as in the gure, with the origin at the point of release. (a) Fill in the constants of Equation 3.14b for the y-displacement, and set the displacement equal to 45.0 m, the y-displacement when the stone hits the ground. Using the quadratic formula, solve for the time. To solve (b), substitute the time from part (a) into the components of the velocity, and substitute the same time into the equation for the x-displacement to solve (c). Solution (a) Find the time of ight. Find the initial x- and y-components of the velocity: v 0x v 0y Find the y-displacement, taking y 0 and v 0y 10.0 m/s: 0, y 45.0 m, v 0 cos v 0 sin y 45.0 m t 4.22 s y 0 0 y (0, 0) v 0 = 20.0 m/s 30.0 x 45.0 m (x, 45.0) x Figure 3.19 (Example 3.7) (20.0 m/s)(cos 30.0 ) (20.0 m/s)(sin 30.0 ) y0 v 0yt 12 2 gt 17.3 m/s 10.0 m/s (10.0 m/s)t (4.90 m/s 2)t 2 Reorganize the equation into standard form and use the quadratic formula (see Appendix A) to nd the positive root: (b) Find the speed at impact. Substitute the value of t found in part (a) into Equation 3.14a to nd the y- component of the velocity at impact: Use this value of v y , the Pythagorean theorem, and the fact that v x v 0x 17.3 m/s to nd the speed of the stone at impact: (c) Find the horizontal range of the stone. Substitute the time of ight into the range equation: vy v 0y gt 10.0 m/s 31.4 m/s (9.80 m/s2)(4.22 s) v vx 2 vy2 (17.3 m/s)2 ( 31.4 m/s)2 35.9 m/s x x x0 (v 0 cos )t (20.0 m/s) (cos 30.0 ) (4.22 s) 73.1 m 3.4 Motion in Two Dimensions 67 Exercise 3.7 Suppose the stone is thrown at an angle of 30.0 degrees below the horizontal. If it strikes the ground 57.0 m away, nd (a) the time of ight, (b) the initial speed, and (c) the speed and the angle of the velocity vector with respect to the horizontal at impact. (Hint: For part (a), use the equation for the x-displacement to eliminate v0t from the equation for the y-displacement.) Answers (a) 1.57 s (b) 41.9 m/s (c) 51.3 m/s, 45.0 Investigate this problem further by logging into PhysicsNow at www.cp7e.com and going to Interactive Example 3.7. So far we have studied only problems in which an object with an initial velocity follows a trajectory determined by the acceleration of gravity alone. In the more general case, other agents, such as air drag, surface friction, or engines, can cause accelerations. These accelerations, taken together, form a vector quantity with components ax and a y . When both components are constant, we can use Equations 3.11 and 3.12 to study the motion, as in the next example. EXAMPLE 3.8 The Rocket Goal Solve a problem involving accelerations in two directions. Problem A jet plane traveling horizontally at 1.00 10 2 m/s drops a rocket from a considerable height. (See Figure 3.20.) The rocket immediately res its engines, accelerating at 20.0 m/s2 in the x-direction while falling under the inuence of gravity in the y-direction. When the rocket has fallen 1.00 km, nd (a) its velocity in the y-direction, (b) its velocity in the x -direction, and (c) the magnitude and direction of its velocity. Neglect air drag and aerodynamic lift. v 0x = 1.00 102 m/s y 1.00 103m Strategy Because the rocket maintains a horizontal orientation (say, through gyroscopes), the x- and y -components of acceleration are independent of each Figure 3.20 (Example 3.8) other. Use the time-independent equation for the velocity in the y-direction to nd the y-component of the velocity after the rocket falls 1.00 km. Then calculate the time of the fall, and use that time to nd the velocity in the x-direction. Solution (a) Find the velocity in the y -direction. Use Equation 3.14c: 9.80 m/s2, and y Substitute v 0y 0, g 3 m, and solve: 1.00 10 (b) Find the velocity in the x-direction. Find the time it takes the rocket to drop 1.00 using the y-component of the velocity: 103 m, 1.40 vx v 0x 10 2 vy m/s v 0y 0 ayt (9.80 m/s 2)t 10 2 m/s : t 14.3 s vy 2 v y2 v 0y 2 0 vy 2g y 2( 9.8 m/s 2)( 1.40 1.00 10 3 m) 10 2 m/s Substitute t, v 0x , and a x into Equation 3.11a to nd the velocity in the x-direction: ax t 1.00 (20.0 m/s 2)(14.3 s) 386 m/s 68 Chapter 3 Vectors and Two-Dimensional Motion (c) Find the magnitude and direction of the velocity. Find the magnitude using the Pythagorean theorem and the results of parts (a) and (b): v v x 2 vy2 ( 1.40 102 m/s)2 (386 m/s)2 411 m/s 1 Use the inverse tangent function to nd the angle: tan vy vx tan 1 1.40 10 2 m/s 386 m/s 19.9 Remarks Notice the symmetry: The kinematic equations for the x- and y-directions are handled in exactly the same way. Having a nonzero acceleration in the x -direction doesnt greatly increase the difculty of the problem. Exercise 3.8 Suppose a rocket-propelled motorcycle is red from rest horizontally across a canyon 1.00 km wide. (a) What minimum constant acceleration in the x -direction must be provided by the engines so the cycle crosses safely if the opposite side is 0.750 km lower than the starting point? (b) At what speed does the motorcycle land if it maintains this constant horizontal component of acceleration? Neglect air drag, but remember that gravity is still acting in the negative y-direction. Answers (a) 13.1 m/s 2 (b) 202 m/s In a stunt similar to that described in Exercise 3.8, motorcycle daredevil Evel Knievel tried to vault across Hells Canyon, part of the Snake River system in Idaho, on his rocket-powered Harley-Davidson X-2 Skycycle. (See the chapter-opening photo on page 53). He lost consciousness at takeoff and released a lever, prematurely deploying his parachute and falling short of the other side. He landed safely in the canyon. 3.5 RELATIVE VELOCITY Relative velocity is all about relating the measurements of two different observers, one moving with respect to the other. The measured velocity of an object depends on the velocity of the observer with respect to the object. On highways, for example, cars moving in the same direction are often moving at high speed relative to Earth, but relative each other they hardly move at all. To an observer at rest at the side of the road, a car might be traveling at 60 mi/h, but to an observer in a truck traveling in the same direction at 50 mi/h, the car would appear to be traveling only 10 mi/h. So measurements of velocity depend on the reference frame of the observer. Reference frames are just coordinate systems. Most of the time, we use a stationary frame of reference relative to Earth, but occasionally we use a moving frame of reference associated with a bus, car, or plane moving with constant velocity relative to Earth. In two dimensions, relative velocity calculations can be confusing, so a systematic approach is important and useful. Let E be an observer, assumed stationary with respect to Earth. Let two cars be labeled A and B, and introduce the following notation (see Figure 3.21): : r AE the position of Car A as measured by E (in a coordinate system xed with respect to Earth). r BE r AB the position of Car B as measured by E. the position of Car A as measured by an observer in Car B. : : According to the preceding notation, the rst letter tells us what the vector is pointing at and the second letter tells us where the position vector starts. The posir r tion vectors of Car A and Car B relative to E, :AE and :BE , are given in the gure. r How do we nd :AB, the position of Car A as measured by an observer in Car B? We simply draw an arrow pointing from Car B to Car A, which can be obtained by r r subtracting :BE from :AE: : r AB : r AE : r BE [3.15] Now, the rate of change of these quantities with time gives us the relationship 3.5 Figure 3.21 The position of Car A relative to Car B can be found by vector subtraction. The rate of change of the resultant vector with respect to time is the relative velocity equation. Relative Velocity 69 y A r AE rAB rAE rBE E rB x E B between the associated velocities: : v AB : v AE : v BE [3.16] The coordinate system of observer E need not be xed to Earth, although it often is. Take careful note of the pattern of subscripts; rather than memorize Equation 3.15, its better to study the short derivation shown in Figure 3.21. Note also that the equation doesnt work for observers traveling a sizeable fraction of the speed of light, when Einsteins theory of special relativity comes into play. PROBLEM-SOLVING STRATEGY Relative Velocity 1. Label each object involved (usually three) with a letter that reminds you of what it is (for example, E for Earth). 2. Look through the problem for phrases such as The velocity of A relative to B, v and write the velocities as :AB. When a velocity is mentioned but it isnt explicitly stated as relative to something, its almost always relative to Earth. 3. Take the three velocities youve found and assemble them into an equation just like Equation 3.16, with subscripts an in analogous order. 4. There will be two unknown components. Solve for them with the x- and y -components of the equation developed in step 3. EXAMPLE 3.9 Pitching Practice on the Train Goal Solve a one-dimensional relative velocity problem. Problem A train is traveling with a speed of 15.0 m/s relative to Earth. A passenger standing at the rear of the train pitches a baseball with a speed of 15.0 m/s relative to the train off the back end, in the direction opposite the motion of the train. What is the velocity of the baseball relative to Earth? Strategy Solving these problems involves putting the proper subscripts on the velocities and arranging them as in Equation 3.16. In the rst sentence of the problem statement, we are informed that the train travels at v 15.0 m/s relative to Earth. This quantity is :TE , with T for train and E for Earth. The passenger throws the basev ball at 15 m/s relative to the train, so this quantity is :BT , where B stands for baseball. The second sentence asks for the velocity of the baseball relative to Earth, :BE. The rest of the problem can be solved by identifying v the correct components of the known quantities and solving for the unknowns, using an analog of Equation 3.16. Solution Write the x-components of the known quantities: (:TE)x v ( v BT)x Follow Equation 3.16: Insert the given values, and solve: (:BT)x v 15 m/s (:BE)x v : 15 m/s 15 m/s (:BE)x v (:BE)x v 0 (:TE)x v 15 m/s 70 Chapter 3 Vectors and Two-Dimensional Motion Exercise 3.9 A train is traveling at 27 m/s relative Earth, and a passenger standing in the train throws a ball at 15 m/s relative the train in the same direction as the trains motion. Find the speed of the ball relative to Earth. Answer 42 m/s EXAMPLE 3.10 Crossing a River Goal Solve a simple two-dimensional relative motion problem. W N E S Problem The boat in Figure 3.22 is heading due north as it crosses a wide river with a velocity of 10.0 km/h relative to the water. The river has a uniform velocity of 5.00 km/h due east. Determine the velocity of the boat with respect to an observer on the riverbank. Strategy Again, we look for key phrases. The boat (has) . . . a velocity of v 10.0 km/h relative to the water gives :BR . The river has a uniform velocity of v 5.00 km/h due east gives :RE , because this implies velocity with respect to Earth. The observer on the riverbank is in a reference frame at rest with respect to Earth. Because were looking for the velocity of the boat with respect to that observer, this last velocity is designated :BE. Take east to be the x-direction, north the v y -direction. Solution Arrange the three quantities into the proper relative velocity equation: Write the velocity vectors in terms of their components. For convenience, these are organized in the following table: vRE vBE vBR u : v BR : v BE : v RE Figure 3.22 (Example 3.10) Vector : x-component (km/h) 0 vx 5.00 y-component (km/h) 10.0 vy 0 v BR : v BE : v RE Find the x -component of velocity: Find the y -component of velocity: Find the magnitude of :BE: v 0 vx 5.00 km/h vy : vx vy 5.00 km/h 10.0 km/h 10.0 km/h vBE 0: vx2 vy2 (5.00 km/h)2 tan 1 (10.0 km/h)2 1 11.2 km/h Find the direction of :BE: v vx vy tan 5.00 m/s 10.0 m/s 26.6 The boat travels at a speed of 11.2 km/h in the direction 26.6 east of north with respect to Earth. Exercise 3.10 Suppose the river is owing east at 3.00 m/s and the boat is traveling south at 4.00 m/s with respect to the river. Find the speed and direction of the boat relative to Earth. Answer 5.00 m/s, 53.1 south of east 3.5 Relative Velocity 71 EXAMPLE 3.11 Bucking the Current Goal Solve a complex two-dimensional relative motion problem. Problem If the skipper of the boat of Example 3.10 moves with the same speed of 10.0 km/h relative to the water, but now wants to travel due north, as in Figure 3.23, in what direction should he head? What is the speed of the boat, according to an observer on the shore? The river is owing east at 5.00 km/h. Strategy Proceed as in the previous problem. In this situation, we must nd the heading of the boat and its velocity with respect to the water, using the fact that the boat travels due north. W N E S W N E S vRE vBE vBR u 60 45 vBE vBR vRE (a) Figure 3.23 (a) (Example 3.11) (b) (Exercise 3.11) : (b) Solution Arrange the three quantities, as before: Organize a table of velocity components: : v BR v BE : v RE y-component (km/h) (10.0 km/h)cos v 0 Vector : x-component (km/h) (10.0 km/h)sin 0 5.00 km/h v BR v BE : v RE : The x-component of the relative velocity equation can be used to nd : (10.0 m/s) sin sin 1.00 2.00 0 5.00 km/h 1.00 2.00 5.00 km/h 10.0 km/h 30.0 Apply the inverse sine function and nd , which is the boats heading, east of north: The y -component of the relative velocity equation can be used to nd v: sin 1 (10.0 km/h)cos v : v 8.66 km/h Remarks From Figure 3.23, we see that this problem can be solved with the Pythagorean theorem, because the problem involves a right triangle: The boats x-component of velocity exactly cancels the rivers velocity. When this is not the case, a more general technique is necessary, as shown in the following exercise. Notice that in the x -component of the relative velocity equation a minus sign had to be included in the term (10.0 km/h) sin because the x -component of the boats velocity with respect to the river is negative. Exercise 3.11 Suppose the river is moving east at 5.00 km/h and the boat is traveling 45.0 south of east with respect to Earth. Find (a) the speed of the boat with respect to Earth and (b) the speed of the boat with respect to the river if the boats heading in the water is 60.0 south of east. (See Figure 3.23b.) You will have to solve two equations and two unknowns. Answers (a) 16.7 km/h (b) 13.7 km/h 72 Chapter 3 Vectors and Two-Dimensional Motion SUMMARY Take a practice test by logging into PhysicsNow at www.cp7e.com and clicking on the Pre-Test link for this chapter. Taking the limit of this expression as t gets arbitrarily a small gives the instantaneous acceleration vector : : : : 3.1 Vectors and their Properties Two vectors A and B can be added geometrically with the triangle method. The two vectors are drawn to scale on graph paper, with the tail of the second vector located at the tip of the rst. The resultant vector is the vector drawn from the tail of the rst vector to the tip of the second. : The negative of a vector A is a vector with the same : magnitude as A , but pointing in the opposite direction. A vector can be multiplied by a scalar, changing its magnitude, and its direction if the scalar is negative. : : a lim t :0 v t [3.10] 3.4 Motion in Two Dimensions The general kinematic equations in two dimensions for objects with constant acceleration are, for the x -direction, vx x vx2 where v 0x v 0 cos 0, v 0x v 0x t v0x2 ax t 1 2 2 axt [3.11a] [3.11b] [3.11c] 2ax x 3.2 Components of a Vector A vector A can be split into two components, one pointing in the x -direction and the other in the y-direction. These components form two sides of a right triangle having a hypotenuse with magnitude A and are given by Ax Ay A cos A sin : : and, for the y-direction, v 0y v0yt v 0y 2 vy y vy 2 a yt 1 2 [3.12a] [3.12b] [3.12c] a yt2 2a y y [3.2] The magnitude and direction of A are related to its components through the Pythagorean theorem and the denition of the tangent: A tan where v 0y v 0 sin 0. The speed v of the object at any instant can be calculated from the components of velocity at that instant using the Pythagorean theorem: v v x2 v y2 A x 2 Ay Ay2 [3.3] [3.4] The angle that the velocity vector makes with the x-axis is given by tan 1 vy vx A If R : R are : : : Ax B , then the components of the resultant vector Rx Ry Ax Ay Bx By [3.5a] [3.5b] The kinematic equations are easily adapted and simplied for projectiles close to the surface of the Earth. The equations for the motion in the horizontal or x-direction are vx x v 0x v 0x t v 0 cos 0 constant 0 )t [3.13a] [3.13b] 3.3 Displacement, Velocity, and Acceleration in Two Dimensions The displacement of an object in two dimensions is dened as the change in the objects position vector: : : : (v 0 cos while the equations for the motion in the vertical or y-direction are vy y vy 2 v0 sin (v0 sin (v 0 sin 0 gt 0)t 2 0) [3.14a] 12 2 gt r rf ri [3.6] [3.14b] [3.14c] The average velocity of an object during the time interval t is : : 2g y v av r t [3.7] t gets arbitrarily Problems are solved by algebraically manipulating one or more of these equations, which often reduces the system to two equations and two unknowns. Taking the limit of this expression as small gives the instantaneous velocity : : v : 3.5 Relative Velocity Let E be an observer, and B a second observer traveling v with velocity :BE as measured by E. If E measures the v velocity of an object A as :AE, then B will measure As velocity as : r : [3.8] lim v t:0 t The direction of the instantaneous velocity vector is along a line that is tangent to the path of the object and in the direction of its motion. The average acceleration of an object with a velocity changing by : in the time interval t is v : : v AB : v AE : v BE [3.16] a av v t [3.9] Solving relative velocity problems involves identifying the velocities properly and labeling them correctly, substituting into Equation 3.16, and then solving for unknown quantities. Problems 73 CONCEPTUAL QUESTIONS 1. Vector A lies in the xy-plane. For what orientations of vec: tor A will both of its components be negative? When will the components have opposite signs? 2. If B is added to A, under what conditions does the resultant vector have a magnitude equal to A B? Under what conditions is the resultant vector equal to zero? 3. A wrench is dropped from the top of a 10-m mast on a sailing ship while the ship is traveling at 20 knots. Where will the wrench hit the deck? (Galileo posed this problem.) 4. Two vectors have unequal magnitudes. Can their sum be zero? Explain. 5. A projectile is red on Earth with some initial velocity. Another projectile is red on the Moon with the same initial velocity. If air resistance is neglected, which projectile has the greater range? Which reaches the greater altitude? (Note that the free-fall acceleration on the Moon is about 1.6 m/s2.) 6. Can a vector A have a component greater than its magnitude A? 7. Is it possible to add a vector quantity to a scalar quantity? 8. Under what circumstances would a vector have components that are equal in magnitude? 9. As a projectile moves in its path, is there any point along the path where the velocity and acceleration vectors are (a) perpendicular to each other? (b) parallel to each other? 10. A rock is dropped at the same instant that a ball is thrown horizontally from the same initial elevation. Which will have the greater speed when it reaches ground level? 11. Explain whether the following particles do or do not have an acceleration: (a) a particle moving in a straight line with constant speed and (b) a particle moving around a curve with constant speed. 12. Correct the following statement: The racing car rounds the turn at a constant velocity of 90 miles per hour. 13. A spacecraft drifts through space at a constant velocity. Suddenly, a gas leak in the side of the spacecraft causes it to constantly accelerate in a direction perpendicular to the initial velocity. The orientation of the spacecraft does not change, so the acceleration remains perpendicular to the original direction of the velocity. What is the shape of the path followed by the spacecraft? : : : : 14. A ball is projected horizontally from the top of a building. One second later, another ball is projected horizontally from the same point with the same velocity. At what point in the motion will the balls be closest to each other? Will the rst ball always be traveling faster than the second? What will be the time difference between them when the balls hit the ground? Can the horizontal projection velocity of the second ball be changed so that the balls arrive at the ground at the same time? 15. Two projectiles are thrown with the same initial speed, one at an angle with respect to the level ground and the other at angle 90 . Both projectiles strike the ground at the same distance from the projection point. Are both projectiles in the air for the same length of time? 16. A baseball is thrown such that its initial x- and ycomponents of velocity are known. Neglecting air resistance, describe how you would calculate the balls (a) coordinates, (b) velocity, and (c) acceleration at the instant the ball reaches the top of its trajectory. How would these results change if air resistance were taken into account? 17. A projectile is red at some angle to the horizontal with some initial speed v0, and air resistance is neglected. Is the projectile a freely falling body? What is its acceleration in the vertical direction? What is its acceleration in the horizontal direction? 18. Determine which of the following moving objects obey the equations of projectile motion developed in this chapter: (a) A ball is thrown in an arbitrary direction. (b) A jet airplane crosses the sky with its engines thrusting the plane forward. (c) A rocket leaves the launch pad. (d) A rocket moves through the sky after its engines have failed. (e) A stone is thrown under water. 19. How can you throw a projectile so that it has zero speed at the top of its trajectory? So that it has nonzero speed at the top of its trajectory? 20. A ball is thrown upward in the air by a passenger on a train that is moving with constant velocity. (a) Describe the path of the ball as seen by the passenger. Describe the path as seen by a stationary observer outside the train. (b) How would these observations change if the train were accelerating along the track? PROBLEMS 1, 2, 3 = straightforward, intermediate, challenging = full solution available in Student Solutions Manual/Study Guide = coached problem with hints available at www.cp7e.com = biomedical application Section 3.1 Vectors and Their Properties 1. A roller coaster moves 200 ft horizontally and then rises 135 ft at an angle of 30.0 above the horizontal. Next, it travels 135 ft at an angle of 40.0 below the horizontal. Use graphical techniques to nd the roller coasters displacement from its starting point to the end of this movement. 2. An airplane ies 200 km due west from city A to city B and then 300 km in the direction of 30.0 north of west from city B to city C. (a) In straight-line distance, how far is city C from city A? (b) Relative to city A, in what direction is city C? 3. A man lost in a maze makes three consecutive displacements so that at the end of his travel he is right back where he started. The rst displacement is 8.00 m westward, and the second is 13.0 m northward. Use the graphical method to nd the magnitude and direction of the third displacement. 74 Chapter 3 Vectors and Two-Dimensional Motion 4. A jogger runs 100 m due west, then changes direction for the second leg of the run. At the end of the run, she is 175 m away from the starting point at an angle of 15.0 north of west. What were the direction and length of her second displacement? Use graphical techniques. 5. A plane ies from base camp to lake A, a distance of 280 km at a direction of 20.0 north of east. After dropping off supplies, the plane ies to lake B, which is 190 km and 30.0 west of north from lake A. Graphically determine the distance and direction from lake B to the base camp. : 6. Vector A has a magnitude of 8.00 units and makes an an: gle of 45.0 with the positive x-axis. Vector B also has a magnitude of 8.00 units and is directed along the negative x-axis. Using graphical methods, nd (a) the vector sum : : : : A B and (b) the vector difference A B. : 7. Vector A is 3.00 units in length and points along the positive : x-axis. Vector B is 4.00 units in length and points along the negative y-axis. Use graphical methods to nd the magni: : : : tude and direction of the vectors (a) A B and (b) A B. : : 8. Each of the displacement vectors A and B shown in Figure : : P3.8 has a magnitude of 3.00 m. Graphically nd (a) A B, : : : : : : (b) A B, (c) B A, and (d) A 2B. y 15. The eye of a hurricane passes over Grand Bahama Island in a direction 60.0 north of west with a speed of 41.0 km/h. Three hours later, the course of the hurricane suddenly shifts due north, and its speed slows to 25.0 km/h. How far from Grand Bahama is the hurricane 4.50 h after it passes over the island? 16. A small map shows Atlanta to be 730 miles in a direction 5 north of east from Dallas. The same map shows that Chicago is 560 miles in a direction 21 west of north from Atlanta. Assume a at Earth, and use the given information to nd the displacement from Dallas to Chicago. 17. A commuter airplane starts from an airport and takes the route shown in Figure P3.17. The plane rst ies to city A, located 175 km away in a direction 30.0 north of east. Next, it ies for 150 km 20.0 west of north, to city B. Finally, the plane ies 190 km due west, to city C. Find the location of city C relative to the location of the starting point. y(km) C 250 200 R 150 100 50 a 30.0 c b B W 20.0 S 110 N E A x (km) B 3.00 m A O 50 100 150 200 Figure P3.17 3.0 O 0m 30.0 x Figure P3.8 18. The helicopter view in Figure P3.18 shows two people pulling on a stubborn mule. Find (a) the single force that is equivalent to the two forces shown and (b) the force that a third person would have to exert on the mule to make the net force equal to zero. The forces are measured in units of newtons (N). Section 3.2 Components of a Vector 9. A golfer takes two putts to get his ball into the hole once he is on the green. The rst putt displaces the ball 6.00 m east, the second 5.40 m south. What displacement would have been needed to get the ball into the hole on the rst putt? 10. A person walks 25.0 north of east for 3.10 km. How far would the person walk due north and due east to arrive at the same location? 11. A girl delivering newspapers covers her route by traveling 3.00 blocks west, 4.00 blocks north, and then 6.00 blocks east. (a) What is her resultant displacement? (b) What is the total distance she travels? 12. While exploring a cave, a spelunker starts at the entrance and moves the following distances: 75.0 m north, 250 m east, 125 m at an angle 30.0 north of east, and 150 m south. Find the resultant displacement from the cave entrance. 13. A vector has an x -component of 25.0 units and a y-component of 40.0 units. Find the magnitude and direction of the vector. 14. A quarterback takes the ball from the line of scrimmage, runs backwards for 10.0 yards, then runs sideways parallel to the line of scrimmage for 15.0 yards. At this point, he throws a 50.0-yard forward pass straight downeld, perpendicular to the line of scrimmage. What is the magnitude of the footballs resultant displacement? y F2 = 80.0 N 75.0 60.0 F1 = 120 N x Figure P3.18 19. A man pushing a mop across a oor causes the mop to undergo two displacements. The rst has a magnitude of 150 cm and makes an angle of 120 with the positive x-axis. The resultant displacement has a magnitude of 140 cm and is directed at an angle of 35.0 to the positive x-axis. Find the magnitude and direction of the second displacement. 20. An airplane starting from airport A ies 300 km east, then 350 km at 30.0 west of north, and then 150 km north to arrive nally at airport B. (a) The next day, another plane Problems 75 B E Joseph Kayne/Dembinsky Photo Associates ies directly from A to B in a straight line. In what direction should the pilot travel in this direct ight? (b) How far will the pilot travel in the ight? Assume there is no wind during either ight. 21. Long John Silver, a pirate, has buried his treasure on an island with ve trees located at the following points: A (30.0 m, 20.0 m), B (60.0 m, 80.0 m), C ( 10.0 m, 10.0 m), D (40.0 m, 30.0 m), and E ( 70.0 m, 60.0 m). All of the points are measured relative to some origin, as in Figure P3.21. Long Johns map instructs you to start at A and move toward B, but cover only one-half the distance between A and B. Then move toward C, covering one-third the distance between your current location and C. Then move toward D, covering one-fourth the distance between where you are and D. Finally, move toward E, covering one-fth the distance between you and E, stop, and dig. (a) What are the coordinates of the point where the pirates treasure is buried? (b) Rearrange the order of the trees for instance, B (30 m, 20 m), A (60 m, 80 m), E ( 10 m, 10 m), C (40 m, 30 m), and D ( 70 m, 60 m) and repeat the calculation to show that the answer does not depend on the order. 23. A peregrine falcon is the fastest bird, ying at a speed of 200 mi/h. Nature has adapted the bird to reach such a speed by placing bafes in its nose to prevent air from rushing in and slowing it down. Also, the birds eyes adjust their focus faster than the eyes of any other creature, so the falcon can focus quickly on its prey. Assume that a peregrine falcon is moving horizontally at its top speed at a height of 100 m above the ground when it brings its wings into its sides and begins to drop in free fall. How far will the bird fall vertically while traveling horizontally a distance of 100 m? Figure P3.23 Notice the structure within the peregrine falcons nostrils. y x C A D Figure P3.21 24. A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18.0 m/s. The cliff is 50.0 m above a at, horizontal beach, as shown in Figure P3.24. How long after being released does the stone strike the beach below the cliff? With what speed and angle of impact does the stone land? y v0 = + 18.0 m/s Section 3.3 Displacement, Velocity, and Acceleration in Two Dimensions Section 3.4 Motion in Two Dimensions 22. One of the fastest recorded pitches in major-league baseball, thrown by Billy Wagner in 2003, was clocked at 101.0 mi/h (Fig. P3.22). If a pitch were thrown horizontally with this velocity, how far would the ball fall vertically by the time it reached home plate, 60.5 ft away? g h = 50.0 m O x v Figure P3.24 Figure P3.22 Billy Wagner throws a baseball. 25. The best leaper in the animal kingdom is the puma, which can jump to a height of 12 ft when leaving the ground at an angle of 45 . With what speed, in SI units, must the animal leave the ground to reach that height? 26. Tom the cat is chasing Jerry the mouse across the surface of a table 1.5 m above the oor. Jerry steps out of the way at the last second, and Tom slides off the edge of the table at a speed of 5.0 m/s. Where will Tom strike the oor, and what velocity components will he have just before he hits? AP/Wide World Photos 76 27. Chapter 3 Vectors and Two-Dimensional Motion 28. 29. 30. 31. 32. 33. 34. A tennis player standing 12.6 m from the net hits the ball at 3.00 above the horizontal. To clear the net, the ball must rise at least 0.330 m. If the ball just clears the net at the apex of its trajectory, how fast was the ball moving when it left the racquet? An artillery shell is red with an initial velocity of 300 m/s at 55.0 above the horizontal. To clear an avalanche, it explodes on a mountainside 42.0 s after ring. What are the x- and y-coordinates of the shell where it explodes, relative to its ring point? A brick is thrown upward from the top of a building at an angle of 25 to the horizontal and with an initial speed of 15 m/s. If the brick is in ight for 3.0 s, how tall is the building? A placekicker must kick a football from a point 36.0 m (about 39 yd) from the goal, and the ball must clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a velocity of 20.0 m/s at an angle of 53 to the horizontal. (a) By how much does the ball clear or fall short of clearing the crossbar? (b) Does the ball approach the crossbar while still rising or while falling? A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24.0 below the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. The car rolls from rest down the incline with a constant acceleration of 4.00 m/s 2 for a distance of 50.0 m to the edge of the cliff, which is 30.0 m above the ocean. Find (a) the cars position relative to the base of the cliff when the car lands in the ocean and (b) the length of time the car is in the air. A reman 50.0 m away from a burning building directs a stream of water from a ground-level re hose at an angle of 30.0 above the horizontal. If the speed of the stream as it leaves the hose is 40.0 m/s, at what height will the stream of water strike the building? A projectile is launched with an initial speed of 60.0 m/s at an angle of 30.0 above the horizontal. The projectile lands on a hillside 4.00 s later. Neglect air friction. (a) What is the projectiles velocity at the highest point of its trajectory? (b) What is the straight-line distance from where the projectile was launched to where it hits its target? A soccer player kicks a rock horizontally off a 40.0-m-high cliff into a pool of water. If the player hears the sound of the splash 3.00 s later, what was the initial speed given to the rock? Assume the speed of sound in air to be 343 m/s. 38. 39. 40. 41. 42. page 109 for an investigation of how the sh can leave the water at a higher speed than it can swim underwater.) If the salmon is in a stream with water speed equal to 1.50 m/s, how high in the air can the sh jump if it leaves the water traveling vertically upwards relative to the Earth? A river ows due east at 1.50 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant velocity of 10.0 m/s due north relative to the water. (a) What is the velocity of the boat relative to the shore? (b) If the river is 300 m wide, how far downstream has the boat moved by the time it reaches the north shore? A rowboat crosses a river with a velocity of 3.30 mi/h at an angle 62.5 north of west relative to the water. The river is 0.505 mi wide and carries an eastward current of 1.25 mi/h. How far upstream is the boat when it reaches the opposite shore? Suppose a Chinook salmon needs to jump a waterfall that is 1.50 m high. If the sh starts from a distance 1.00 m from the base of the ledge over which the waterfall ows, nd the x- and y-components of the initial velocity the salmon would need to just reach the ledge at the top of its trajectory. Can the sh make this jump? (Remember that a Chinook salmon can jump out of the water with a speed of 6.26 m/s.) How long does it take an automobile traveling in the left lane of a highway at 60.0 km/h to overtake (become even with) another car that is traveling in the right lane at 40.0 km/h when the cars front bumpers are initially 100 m apart? A science student is riding on a atcar of a train traveling along a straight horizontal track at a constant speed of 10.0 m/s. The student throws a ball along a path that she judges to make an initial angle of 60.0 with the horizontal and to be in line with the track. The students professor, who is standing on the ground nearby, observes the ball to rise vertically. How high does the ball rise? ADDITIONAL PROBLEMS 43. A particle undergoes two displacements. The rst has a magnitude of 150 cm and makes an angle of 120.0 with the positive x-axis. The resultant of the two displacements is 140 cm, directed at an angle of 35.0 to the positive x-axis. Find the magnitude and direction of the second displacement. 44. Find the sum of these four vector forces: 12.0 N to the right at 35.0 above the horizontal, 31.0 N to the left at 55.0 above the horizontal, 8.40 N to the left at 35.0 below the horizontal, and 24.0 N to the right at 55.0 below the horizontal. (Hint: N stands for newton, the SI unit of force. The component method allows the addition of any vectors forces as well as displacements and velocities. Make a drawing of this situation, and select the best axes for x and y so that you have the least number of components.) 45. A car travels due east with a speed of 50.0 km/h. Rain is falling vertically with respect to Earth. The traces of the rain on the side windows of the car make an angle of 60.0 with the vertical. Find the velocity of the rain with respect to (a) the car and (b) Earth. 46. You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the Section 3.5 Relative Velocity 35. A jet airliner moving initially at 300 mi/h due east enters a region where the wind is blowing at 100 mi/h in a direction 30.0 north of east. What is the new velocity of the aircraft relative to the ground? 36. A boat moves through the water of a river at 10 m/s relative to the water, regardless of the boats direction. If the water in the river is owing at 1.5 m/s, how long does it take the boat to make a round trip consisting of a 300-m displacement downstream followed by a 300-m displacement upstream? 37. A Chinook (King) salmon (Genus Oncorynchus) can jump out of water with a speed of 6.26 m/s. (See Problem 4.9, Problems 77 truth of this statement, consider a ball thrown off the top of a building with a velocity : at an angle with respect to v the horizontal. Let the building be 50.0 m tall, the initial horizontal velocity be 9.00 m/s, and the initial vertical velocity be 12.0 m/s. Choose your coordinates such that the positive y-axis is upward, the x-axis is to the right, and the origin is at the point where the ball is released. (a) With these choices, nd the balls maximum height above the ground, and the time it takes to reach the maximum height. (b) Repeat your calculations choosing the origin at the base of the building. 47. Towns A and B in Figure P3.47 are 80.0 km apart. A couple arranges to drive from town A and meet a couple driving from town B at the lake, L. The two couples leave simultaneously and drive for 2.50 h in the directions shown. Car 1 has a speed of 90.0 km/h. If the cars arrive simultaneously at the lake, what is the speed of car 2? 51. 52. 53. L 1 2 54. A 40.0 80.0 km B Figure P3.47 48. A Chinook salmon has a maximum underwater speed of 3.58 m/s, but it can jump out of water with a speed of 6.26 m/s. To move upstream past a waterfall, the salmon does not need to jump to the top of the fall, but only to a point in the fall where the water speed is less than 3.58 m/s; it can then swim up the fall for the remaining distance. Because the salmon must make forward progress in the water, lets assume that it can swim to the top if the water speed is 3.00 m/s. If water has a speed of 1.50 m/s as it passes over a ledge, how far below the ledge will the water be moving with a speed of 3.00 m/s? (Note that water undergoes projectile motion once it leaves the ledge.) If the salmon is able to jump vertically upward from the base of the fall, what is the maximum height of waterfall that the salmon can clear? 49. A rocket is launched at an angle of 53.0 above the horizontal with an initial speed of 100 m/s. The rocket moves for 3.00 s along its initial line of motion with an acceleration of 30.0 m/s2. At this time, its engines fail and the rocket proceeds to move as a projectile. Find (a) the maximum altitude reached by the rocket, (b) its total time of ight, and (c) its horizontal range. 50. Two canoeists in identical canoes exert the same effort paddling and hence maintain the same speed relative to the water. One paddles directly upstream (and moves upstream), whereas the other paddles directly downstream. With downstream as the positive direction, an observer on shore determines the velocities of the two canoes to be 1.2 m/s and 2.9 m/s, respectively. (a) What is the speed of the water relative to the shore? (b) What is the speed of each canoe relative to the water? If a person can jump a maximum horizontal distance (by using a 45 projection angle) of 3.0 m on Earth, what would be his maximum range on the Moon, where the free-fall acceleration is g/6 and g 9.80 m/s2? Repeat for Mars, where the acceleration due to gravity is 0.38g. A daredevil decides to jump a canyon. Its walls are equally high and 10 m apart. He takes off by driving a motorcycle up a short ramp sloped at an angle of 15 . What minimum speed must he have in order to clear the canyon? : (a) Vector A is in the rst quadrant of a Cartesian coordi: nate system. What is the sign of the x-component of A? : What is the sign of the y-component? (b) Vector B is in the second quadrant of a Cartesian coordinate system. : What is the sign of the x-component of B? What is the : : sign of the y-component? (c) The vector sum A B . Choose the correct ll-in-the-blank answer from (i) must be in either the rst or the second quadrant or : (ii) could be in any quadrant. (d) Let A 30 m at an an: gle of 30 from the positive x-axis and B 20 m at an angle of 40 from the negative x-axis. Test your predictions of parts (a) through (c). A boy and a girl are tossing an apple back and forth between them. Figure P3.54 shows one path the apple follows when watched by an observer looking on from the side. The apple is moving from left to right. Five points are marked on the path. Ignore air resistance. (a) Make a copy of this gure. At each of the marked points, draw an arrow that indicates the magnitude and direction of the apples velocity when it passes through that point. (b) Make a second copy of the gure. This time, at each marked point, place an arrow indicating the magnitude and direction of any acceleration the apple exhibits at that point. Figure P3.54 55. A home run is hit in such a way that the baseball just clears a wall 21 m high, located 130 m from home plate. The ball is hit at an angle of 35 to the horizontal, and air resistance is negligible. Find (a) the initial speed of the ball, (b) the time it takes the ball to reach the wall, and (c) the velocity components and the speed of the ball when it reaches the wall. (Assume that the ball is hit at a height of 1.0 m above the ground.) 56. A ball is thrown straight upward and returns to the throwers hand after 3.00 s in the air. A second ball is thrown at an angle of 30.0 with the horizontal. At what speed must the second ball be thrown so that it reaches the same height as the one thrown vertically? 57. A quarterback throws a football toward a receiver with an initial speed of 20 m/s at an angle of 30 above the horizontal. At that instant, the receiver is 20 m from the 78 Chapter 3 Vectors and Two-Dimensional Motion quarterback. In what direction and with what constant speed should the receiver run in order to catch the football at the level at which it was thrown? 58. A 2.00-m-tall basketball player is standing on the oor 10.0 m from the basket, as in Figure P3.58. If he shoots the ball at a 40.0 angle with the horizontal, at what initial speed must he throw the basketball so that it goes through the hoop without striking the backboard? The height of the basket is 3.05 m. 200 cm, and re-form the vector sums as in part (a). Then nd the vector difference between the two sums. d 2m 23.0 d 2f 28.0 d1m d1f 40.0 Figure P3.60 3.05 m 2.00 m 10.0 m Figure P3.58 59. In a very popular lecture demonstration, a projectile is red at a falling target as in Figure P3.59. The projectile leaves the gun at the same instant that the target is dropped from rest. Assuming that the gun is initially aimed at the target, show that the projectile will hit the target. (One restriction of this experiment is that the projectile must reach the target before the target strikes the oor.) 61. By throwing a ball at an angle of 45 , a girl can throw the ball a maximum horizontal distance R on a level eld. How far can she throw the same ball vertically upward? Assume that her muscles give the ball the same speed in each case. (Is this assumption valid?) 62. A projectile is red with an initial speed v0 at an angle 0 to the horizontal, as in Figure 3.14. When it reaches its peak, the projectile has (x, y) coordinates given by (R/2, h), and when it strikes the ground, its coordinates are (R, 0), where R is called the horizontal range. (a) Show that the projectile reaches a maximum height given by h v 02 sin2 2g 0 Target (b) Show that the horizontal range of the projectile is given by R v 02 sin 2 g 0 v0 0 Point of collision Figure P3.59 60. Figure P3.60 illustrates the difference in proportions between the male (m) and female (f) anatomies. The dis: : placements d 1m and d 1f from the bottom of the feet to the navel have magnitudes of 104 cm and 84.0 cm, respec: : tively. The displacements d 2m and d 2f have magnitudes of 50.0 cm and 43.0 cm, respectively. (a) Find the vector sum : : of the displacements d 1 and d 2 in each case. (b) The male gure is 180 cm tall, the female 168 cm. Normalize the displacements of each gure to a common height of 63. A hunter wishes to cross a river that is 1.5 km wide and ows with a speed of 5.0 km/h parallel to its banks. The hunter uses a small powerboat that moves at a maximum speed of 12 km/h with respect to the water. What is the minimum time necessary for crossing? 64. A water insect maintains an average position on the surface of a stream by darting upstream (against the current) then drifting downstream (with the current) to its original position. The current in the stream is 0.500 m/s relative to the shore, and the insect darts upstream 0.560 m (relative to a spot on shore) in 0.800 s during the rst part of its motion. Take upstream as the positive direction. (a) Determine the velocity of the insect relative to the water (i) during its dash upstream and (ii) during its drift downstream. (b) How far upstream relative to the water does the insect move during one cycle of its motion? (c) What is the average velocity of the insect relative to the water? A daredevil is shot out of a cannon at 65. 45.0 to the horizontal with an initial speed of 25.0 m/s. A net is positioned a horizontal distance of 50.0 m from the cannon. At what height above the cannon should the net be placed in order to catch the daredevil? Problems 79 66. Chinook salmon are able to move upstream faster by jumping out of the water periodically; this behavior is called porpoising. Suppose a salmon swimming in still water jumps out of the water with a speed of 6.26 m/s at an angle of 45 , sails through the air a distance L before returning to the water, and then swims a distance L underwater at a speed of 3.58 m/s before beginning another porpoising maneuver. Determine the average speed of the sh. 67. A student decides to measure the muzzle velocity of a pellet shot from his gun. He points the gun horizontally. He places a target on a vertical wall a distance x away from the gun. The pellet hits the target a vertical distance y below the gun. (a) Show that the position of the pellet when traveling through the air is given by y Ax 2, where A is a constant. (b) Express the constant A in terms of the initial (muzzle) velocity and the free-fall acceleration. (c) If x 3.00 m and y 0.210 m, what is the initial speed of the pellet? 68. A sailboat is heading directly north at a speed of 20 knots (1 knot 0.514 m/s). The wind is blowing towards the east with a speed of 17 knots. Determine the magnitude and direction of the wind velocity as measured on the boat. What is the component of the wind velocity in the direction parallel to the motion of the boat? (See Problem 4.54 for an explanation of how a sailboat can move into the wind.) 69. Instructions for nding a buried treasure include the following: Go 75 paces at 240 , turn to 135 and walk 125 paces, and then travel 100 paces at 160 . Determine the resultant displacement from the starting point. 70. When baseball outelders throw the ball, they usually allow it to take one bounce, on the theory that the ball arrives at its target sooner that way. Suppose that, after the bounce, the ball rebounds at the same angle that it had when it was released (as in Fig. P3.70), but loses half its speed. (a) Assuming that the ball is always thrown with the same initial speed, at what angle should the ball be thrown in order to go the same d istance D with one bounce as a ball thrown upward at 45.0 with no bounce? (b) Determine the ratio of the times for the one-bounce and no-bounce throws. as the rst? (b) How many seconds later should the second snowball be thrown after the rst in order for both to arrive at the same time? 72. A dart gun is red while being held horizontally at a height of 1.00 m above ground level and while it is at rest relative to the ground. The dart from the gun travels a horizontal distance of 5.00 m. A college student holds the same gun in a horizontal position while sliding down a 45.0 incline at a constant speed of 2.00 m/s. How far will the dart travel if the student res the gun when it is 1.00 m above the ground? 73. The determined Wile E. Coyote is out once more to try to capture the elusive roadrunner. The coyote wears a new pair of Acme power roller skates, which provide a constant horizontal acceleration of 15 m/s2, as shown in Figure P3.73. The coyote starts off at rest 70 m from the edge of a cliff at the instant the roadrunner zips by in the direction of the cliff. (a) If the roadrunner moves with constant speed, nd the minimum speed the roadrunner must have in order to reach the cliff before the coyote. (b) If the cliff is 100 m above the base of a canyon, nd where the coyote lands in the canyon. (Assume that his skates are still in operation when he is in ight and that his horizontal component of acceleration remains constant at 15 m/ 2.) Coyote Roadrunner stupidus delightus EP BE BEE P Figure P3.73 u 45.0 D Figure P3.70 u 71. One strategy in a snowball ght is to throw a snowball at a high angle over level ground. Then, while your opponent is watching that snowball, you throw a second one at a low angle timed to arrive before or at the same time as the rst one. Assume that both snowballs are thrown with a speed of 25.0 m/s. The rst is thrown at an angle of 70.0 with respect to the horizontal. (a) At what angle should the second snowball be thrown to arrive at the same point ACTIVITIES A.1. Take three steps, turn 90 , and then walk four steps. Now count the number of steps it takes to walk in a straight line back to your starting point. Verify your result mathematically. A.2. For this investigation, you need to be outside with a small ball such as a tennis ball and a wristwatch with a second hand. Throw the ball vertically upward as hard as you can, and nd the initial speed of your throw and the approximate maximum height of the ball solely with the use of your wristwatch. What happens when you throw the ball at some angle other than 90 ? Does this change the time of ight? Can you still determine the maximum height and initial speed? Give careful explanations for your answers. Your co-worker can eyeball the maximum height by standing at a distance and noting the angle. A.3. Using as projectiles the drops of water spraying from a garden hose at ground level, test the statement that the maximum range occurs when the angle of projection is 45 . As an additional part of this experiment, hold the hose horizontally above the ground and have your coworker position a marker at the location where the water 80 Chapter 3 Vectors and Two-Dimensional Motion strikes the ground. Increase the angle of inclination by about 10 , and record the strike position again. Repeat until you have reached an angle of about 75 . Note the pattern produced. Are there two angles at which the range is the same? Explain the reasoning behind your observations. A.4. Use a chessboard as a coordinate system, with the intersection of the lines on the board as the positions of the coordinates. Select an origin for your coordinate system and, on index cards, write down several vector displacements that are at right angles to one another. For example, displacement 1 might be a movement of four units to the right, displacement 2 an upward movement of six units, and so forth. Continue this for a total of seven or eight movements until you end up at some particular location on the chessboard. Let your co-worker start at the origin and follow your vector directions to see whether he or she arrives at the expected nal location. Now shufe the cards and repeat the experiment. Does the order of the displacements make any difference as to where you eventually end up? A.5. Roll a ball off a table. At the very instant the rolling ball leaves the table, drop a second ball from the same height above the oor. (Doing this will require a sharp eye and good reexes!) Do the two balls hit the oor at the same time? Try varying the speed at which you roll the ball off the table. Does this change affect the time at which the balls strike the oor? Finally, roll one of the balls down an incline, and drop the other ball from the base of the incline at the instant the rst ball leaves the slope. Which of these balls hits the oor rst in this situation? Explain the reasoning behind your observations. ... View Full Document

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